Question: Divide the following complex numbers. $ \dfrac{-11-3i}{1-2i}$
We can divide complex numbers by multiplying both numerator and denominator by the denominator's complex conjugate , which is ${1+2i}$ $ \dfrac{-11-3i}{1-2i} = \dfrac{-11-3i}{1-2i} \cdot \dfrac{{1+2i}}{{1+2i}} $ We can simplify the denominator using the fact $(a + b) \cdot (a - b) = a^2 - b^2$ $ \dfrac{(-11-3i) \cdot (1+2i)} {(1-2i) \cdot (1+2i)} = \dfrac{(-11-3i) \cdot (1+2i)} {1^2 - (-2i)^2} $ Evaluate the squares in the denominator and subtract them. $ \dfrac{(-11-3i) \cdot (1+2i)} {(1)^2 - (-2i)^2} = $ $ \dfrac{(-11-3i) \cdot (1+2i)} {1 + 4} = $ $ \dfrac{(-11-3i) \cdot (1+2i)} {5} $ Note that the denominator now doesn't contain any imaginary unit multiples, so it is a real number, simplifying the problem to complex number multiplication. Now, we can multiply out the two factors in the numerator. $ \dfrac{({-11-3i}) \cdot ({1+2i})} {5} = $ $ \dfrac{{-11} \cdot {1} + {-3} \cdot {1 i} + {-11} \cdot {2 i} + {-3} \cdot {2 i^2}} {5} $ Evaluate each product of two numbers. $ \dfrac{-11 - 3i - 22i - 6 i^2} {5} $ Finally, simplify the fraction. $ \dfrac{-11 - 3i - 22i + 6} {5} = \dfrac{-5 - 25i} {5} = -1-5i $